Improving Spark Performance With Partitioning

At Sortable we use Spark for many of our data processing tasks. Spark is an engine that is scalable: it allows us to run many tasks in parallel across hundreds of machines in a cluster, and can also be used to run tasks across cores on a desktop.

The main abstraction in Spark is the Resilient Distributed Dataset (RDD). RDDs are collections of objects. Under the hood, these objects are stored in partitions. When performing computations on RDDs, these partitions can be operated on in parallel.

One of the tricks that we’ve found to improve the performance of Spark jobs is to change the partitioning of our data. We’ll illustrate how this can help with a simple example.

Finding Prime Numbers

Let’s say we want to find all the prime numbers up to 2 million. We’ll do this very naively by first finding all composite (non-prime) numbers, then find all numbers that are not composite to get the prime numbers.

We’ll find all composite numbers by taking every natural number from 2 to 2 million, then find all multiples of these numbers that are smaller than or equal to 2 million. We will have many duplicates (6 will show up in our multiples of both 2 and 3, for example), but that’s ok.

Let’s do this in the Spark shell:

Welcome to
      ____              __
     / __/__  ___ _____/ /__
    _ / _ / _ `/ __/  '_/
   /___/ .__/_,_/_/ /_/_   version 1.6.1

Using Scala version 2.10.5 (OpenJDK 64-Bit Server VM, Java 1.7.0_101)
Type in expressions to have them evaluated.
Type :help for more information.
Spark context available as sc.
SQL context available as sqlContext.

scala> val n = 2000000
n: Int = 2000000

scala> val composite = sc.parallelize(2 to n, 8).map(x => (x, (2 to (n / x)))).flatMap(kv => * kv._1))
composite: org.apache.spark.rdd.RDD[Int] = MapPartitionsRDD[2] at flatMap at :29

scala> val prime = sc.parallelize(2 to n, 8).subtract(composite)
prime: org.apache.spark.rdd.RDD[Int] = MapPartitionsRDD[7] at subtract at :31

scala> prime.collect()
res0: Array[Int] = Array(563249, 17, 281609, 840761, 1126513, 1958993, 840713, 1959017, 41, 281641, 1681513, 1126441, 73, 1126457, 89, 840817, 97, 1408009, 113, 137, 1408241, 563377, 1126649, 281737, 281777, 840841, 1408217, 1681649, 281761, 1408201, 1959161, 1408177, 840929, 563449, 1126561, 193, 1126577, 1126537, 1959073, 563417, 233, 281849, 1126553, 563401, 281833, 241, 563489, 281, 281857, 257, 1959241, 313, 841081, 337, 1408289, 563561, 281921, 353, 1681721, 409, 281993, 401, 1126897, 282001, 1126889, 1959361, 1681873, 563593, 433, 841097, 1959401, 1408417, 1959313, 1681817, 457, 841193, 449, 563657, 282089, 282097, 1408409, 1408601, 1959521, 1682017, 841241, 1408577, 569, 1408633, 521, 841273, 1127033, 841289, 617, 1408529, 1959457, 563777, 841297, 1959473, 577, 593, 563809, 601,...

The answer looks reasonable, but let’s look at the program’s performance. If we go into the Spark UI we can see that Spark used 3 stages. Here’s the DAG (Directed Acyclic Graph) visualization from the UI, which shows the various RDDs that were computed along the way:

A new stage begins every time the job requires communication between partitions (in Spark terminology this communication is known as a “shuffle”). Spark stages have one task for every partition, and these tasks transform a partition from one RDD into a partition of another RDD. Let’s briefly look at Stage 0’s tasks:

The columns of interest to us are “Duration” and “Shuffle Write Size / Records”. Our sc.parallelize(2 to n, 8) operation created 1999999 records, which were evenly distributed across the 8 partitions. Each task took about the same amount of time, so this looks good.

Stage 1 is the most interesting stage, since it ran our map and flatMap transformations. Let’s look at it too:

This doesn’t look very good, because the work load was not well balanced across the tasks. Over 93% of the data ended up in a single partition, whose task took 14 seconds to run. The next slowest task took 1 second. Since we had 8 cores dedicated to this job, this means that we had 7 idle cores for 13 seconds while they waited for the stage to complete. This is a very inefficient use of resources.

Why did this happen?

When we ran sc.parallelize(2 to n, 8) , Spark used a partitioning scheme that nicely divided the data into 8 even groups. It most likely used a range partitioner, where the numbers from 2-250000 were in the first partition, 250001-500000 in the second, etc. However, our map turned this into (key,value) pairs where the values had wildly different sizes. Each value was a list of all integers we needed to multiply the key by to find the multiples up to 2 million. For half of them (all keys greater than 1 million) this meant that the value was an empty list. Our largest value was for key 2, which had all integers from 2 to 1000000. This is why the first partition had most of the data and took the greatest amount of time, while the last four had no data.

How do we fix this?

We can repartition our data. Calling .repartition(numPartitions) on an RDD in Spark will shuffle the data into the number of partitions we specify. Let’s try adding that to our code.

We’ll run the same thing as before, but insert .repartition(8) between the .map and the .flatMap . Our RDD will have the same number of partitions as before, but the data will be redistributed across those partitions. Then our second line looks like this:

val composite = sc.parallelize(2 to n, 8).map(x => (x, (2 to (n / x)))).repartition(8).flatMap(kv => * kv._1))

Our new graph is a little more complicated because the repartition added another shuffle:

Stage 0 is the same as before. Our new Stage 1 looks very similar to Stage 0, with each task having about 250000 records and taking about 1 second. The interesting one is Stage 2:

This looks much better than our old Stage 1. We still process the same number of records, but this time the work is more evenly distributed across the 8 partitions. Each task took about 5 seconds and our cores were used more efficiently.

In both versions of our code, the final stage took about 6 seconds. So the total time for our first version was about 0.5 + 14 + 6 = ~21 seconds, and with our changes the second one took 0.5 + 1 + 5 + 6 = ~13 seconds. The second version had to do some extra work to repartition the data, but that modification reduced the overall time because it allowed us to use our resources more efficiently.

Of course, if the goal is to find prime numbers, then there are much more efficient algorithms out there. But this example demonstrates the importance of considering how our Spark data is distributed. Introducing a .repartition will increase the amount of work that the Spark engine has to do, but the benefit can significantly outweigh that cost. If we’re clever, we could even skip the .repartition by partitioning the initial data source in a better way.

稿源:Sortable Engineering Blog (源链) | 关于 | 阅读提示

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